Q. One mole of ethanol is produced reacting graphite, $H_{2}$ and $O_{2}$ together. The standard enthalpy of formation is $- \, \text{277} \text{.7} \, \text{kJ} \, \text{mol}^{- \text{1}}$ . Calculate the standard enthalpy of the reaction when 4 moles of graphite is involved.

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A

$- \, 277.7 \, \text{kJ/mol}$

B

$- \, 555.4 \, \text{kJ/mol}$

C

$- \, 138.85 \, \text{kJ/mol}$

D

$- \, 69.42 \, \text{kJ/mol}$

Solution:

$\text{2} \, \text{C(s)} \, \text{+} \, \frac{\text{1}}{\text{2}} \left(\text{O}\right)_{\text{2}} \text{(g)} \, \text{+} \, \text{3} \, \left(\text{H}\right)_{\text{2}} \text{(g)} \rightarrow \left(\text{C}\right)_{\text{2}} \left(\text{H}\right)_{\text{5}} \text{OH(l);} \, \, \left(\text{ΔH}\right)_{\text{f}} \, \text{=} \, - \text{277} \text{.7} \, \text{kJ}$
By doubling the reaction,
$\text{4} \, \text{C(s)} \, \text{+} \, \left(\text{O}\right)_{\text{2}} \text{(g)} \, \text{+} \, \text{6} \, \left(\text{H}\right)_{\text{2}} \text{(g)} \rightarrow \text{2} \, \left(\text{C}\right)_{\text{2}} \left(\text{H}\right)_{\text{5}} \text{OH;} \, \, \left(\text{ΔH}\right)_{\text{f}} \, \text{=} \, \text{2} \, \text{\times } \text{} \, - \text{277} \text{.7}$
$= - \, 555.4 \, \text{kJ}$

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